reactjs - 未捕获的类型错误：_this.props.saveApp(...).then 不是函数

Question

// Component code
this.props.saveApp(values, data).then((result) => {
          //do something

}).catch((error) => {

});

我正在使用反应还原

以下是我的行动

https://pastebin.com/embed_js/EA98HRpj

标签： reactjsreact-nativereact-redux

Answer 1

您需要在您的操作中返回承诺，并且响应未在正确的位置返回。查看我的评论以了解我编辑的内容：

export function saveApp(appvalue, fdata) {

    return dispatch => {

        return request // You need to add the return here!!
          .post(\'http://appplay.net:8082/api/v1/addFile\')
          .send(fdata)
          .then(function(res) {
            // res.body, res.headers, res.status
            console.log(res.body.data);
            var downloadurl = res.body.data;
            // console.log(values);
            // this.saveApp( values, downloadurl );
            var platfrom1 = (appvalue.Platform === "iOS")? "2" : "1";
            var ref_newapp = db.ref().child("Apps");
            var auto_id = ref_newapp.push();
            auto_id.set({
                "App_Icon": appvalue.appicon,
                "App_Name": appvalue.appname,
                "Download_URL": downloadurl,
                "Is_Active": parseInt(appvalue.radiostatus, 10),
                "Platform": parseInt(platfrom1, 10),
                "Version" :appvalue.version
            });

            return res; // I think you want to return the res here!!
          })
          .catch(function(err) {
            // err.message, err.response
          });
    }
}